The common tangents to the circle x2+y2=2 and the parabola y2=8x touch the circle at the points P,Q and the parabola at the points R,S. Then the area of the quadrilateral PQRS is [2014]
(4)
Let the tangent to y2=8x be y=mx+2m
If it is common tangent to parabola and circle x2+y2=2, then distance of the tangent from the centre of the circle is equal to radius of the circle
∴ |2mm2+1|=2⇒4m2(1+m2)=2
⇒m4+m2-2=0 ⇒(m2+2)(m2-1)=0 ⇒m=1 or -1
∴ Required tangents are y=x+2 and y=-x-2
Their common point is (-2,0)
∴ Tangents are drawn from (-2,0)
∴ Chord of contact PQ to circle is x.(-2)+y.(0)=2⇒x=-1
and chord of contact RS to parabola is y(0)=4(x-2)⇒x=2
Hence coordinates of P and Q are (-1,1) and (-1,-1) respectively.
Also coordinates of R and S are (2,-4) and (2,4) respectively.
∴ Area of trapezium PQRS=12(2+8)×3=15